NCERT Exemplar Solutions Class 10 Maths Chapter 7 Coordinate Geometry

Read it straight off the y-value.

• Why: the height of a point above the x-axis is exactly its y-coordinate, so P(2,3) sits three units up and the answer is 3.
• The decoy: the x-coordinate only fixes the left to right position, so the value 2 never enters this distance at all.
• Safe habit: name the axis first, since the x-axis means y=0, and then read the matching coordinate rather than grabbing the first number you see.

Option (B), 3.

A vertical segment, nothing more.

• Setup: both points have x=0, so A and B sit on the y-axis and the line joining them is vertical.
• Shortcut: the length of a vertical segment is just the gap between the two heights, here 6-(-2)=8.
• Cross-check: the full distance formula returns the same √64=8, so both routes agree and the quick one is safe to trust.

Option (B), 8.

A 6-8-10 right triangle in disguise.

• Picture: dropping perpendiculars from P(-6,8) to the two axes builds a right triangle whose legs are 6 and 8.
• Hypotenuse: the line back to the origin is that triangle's hypotenuse, √6²+8²=10, the famous 6-8-10 set.
• Payoff: once you recognise such Pythagorean triples you can write the answer instantly, with no calculator needed.

Option (C), 10.

Two equal legs give a √2 answer.

• Equal gaps: the horizontal gap and the vertical gap are both 5, so the two points are corners of a right isosceles triangle.
• Standard form: the hypotenuse of such a triangle is always a leg times √2, which here gives 5√2 at once.
• Why bother: spotting the equal legs is faster than grinding out √50 and it stops you leaving the surd unsimplified.

Option (B), 5√2.

Both diagonals of a rectangle are equal.

• Sides: the two sides meeting at O are 3 and 5, and they are perpendicular by construction.
• Diagonal: the diagonal across the rectangle is therefore √3²+5²=√34, the length we want.
• Check: the other diagonal OC has the same length, a quick consistency test, and the only care needed is reading A and B as opposite corners rather than adjacent ones.

Option (C), √34.

A right triangle sitting on the axes.

• Legs: two sides run along the axes with lengths 3 and 4, so they are the legs of a right angle at the origin.
• Slant side: the third side is the 3-4-5 hypotenuse, namely 5, so the perimeter is 3+4+5=12.
• The trap: the decoy 7+5 comes from mis-squaring the hypotenuse, so compute √9+16 carefully and that error disappears.

Option (B), 12.

Use the flat base.

• Base: because A(3,0) and B(7,0) share y=0, the base AB runs along the x-axis with length 7-3=4.
• Height: the height is simply how high C rises above that axis, which is 4, so the area is 1244=8.
• When to use it: the vertex formula confirms the same 8, but this base-times-height idea is faster whenever one side is horizontal or vertical.

Option (C), 8.

Symmetry hints at isosceles.

• Mirror: the base endpoints (-4,0) and (4,0) are mirror images across the y-axis, and the apex (0,3) sits right on that mirror line.
• Equal sides: so the two slant sides must come out equal, each 5 by the distance formula, while the base measures 8.
• Final read: two equal sides with one different make it isosceles, and the failed Pythagoras test (25+2564) rules out any right angle.

Option (B), isosceles triangle.

Care with which weight goes where.

• Where it sits: in the ratio 1:2 the part nearer A is smaller, so the point lands closer to A(7,-6) and gets dragged down into negative y.
• The point: the formula gives (173,-83), with positive x and negative y, which is the fourth quadrant.
• Common slip: students swap the cross-multiplication, so write m₁x₂+m₂x₁m₁+m₂ out in full before substituting.

Option (D), IV quadrant.

The midpoint is the safe pick.

• Key fact: the midpoint of a segment always lies on its perpendicular bisector, and here that midpoint works out to the origin (0,0).
• Confirm: a single distance check seals it, since OA=OB=√29 are equal.
• Why faster: testing each other option would mean two distance computations apiece, so starting from the midpoint skips all of that and lands on the answer at once.

Option (A), (0,0).

Equal midpoints pin down D.

• Centre: the diagonals share a midpoint, found from A and C to be (3,3).
• Solve: force the midpoint of B(6,7) and D to be that same point and the fourth vertex falls out as D=(0,-1).
• Neat check: in a parallelogram the step A→ B matches the step D→ C, and indeed both equal (8,4), which confirms the vertex.

Option (B), (0,-1).

Compare in units of 5.

• Common unit: all three lengths come out as multiples of 5, namely AP=5, AB=25 and PB=35.
• Read off: once they share that common factor the relations are easy to compare, and AP=12AB is the one that holds.
• Why it helps: working in the single surd unit removes the clutter of separate square roots and makes the matching option obvious.

Option (D), AP=12 AB.

The y-value is a free check.

• Midpoint: averaging the coordinates of Q and R gives the midpoint (-4,4).
• Real work: the given point already shows y=4, matching perfectly, so the only equation left is a3=-4, giving a=-12.
• Reassurance: that built-in agreement on the y-coordinate tells you no arithmetic slipped before you solved for a.

Option (B), a=-12.

Equidistance on the axis.

• Form: any point on the y-axis is (0,y), so only one unknown is in play.
• Linear: demanding equal distance to A and B drops the y² terms, and solving 26-10y=52-12y gives y=13, so the crossing point is (0,13).
• Decoy: the squared-distance form is the clean route in, and the x-axis option (13,0) is a trap because we wanted a point on the y-axis.

Option (A), (0,13).

Hypotenuse midpoint is the centre.

• Diameter: with the right angle at O, the hypotenuse AB becomes a diameter of the circle through all three vertices.
• Centre: so the midpoint of AB is equidistant from A, O and B, and averaging A(0,2y) with B(2x,0) gives (x,y).
• Why those numbers: the factor-of-two coordinates in the figure are chosen so the midpoint comes out as the clean (x,y), which is exactly option (A).

Option (A), (x,y).

Compare squared distances with r².

• No surds: squaring keeps everything tidy, with the squared radius at r²=42.25.
• Inside: three of the points give squared distances well below that (1.56, 9.44 and 25.25), so they sit in the interior.
• The odd one: the fourth point gives exactly 36+6.25=42.25, equal to r², so it rests on the circle, and ``interior'' needs a strict inequality, which is why it fails.

Option (D), (-6,52).

Double the midpoint, place the zeros.

• Idea: when one endpoint contributes a zero, each coordinate of the other endpoint is just twice the midpoint's matching coordinate.
• Apply: doubling -5 gives the y-axis point (0,-10), and doubling 2 gives the x-axis point (4,0).
• Where zeros come from: each point lies on its own axis, so the answer drops out without solving any equation.

Option (D), (0,-10) and (4,0).

The symmetric form forces zero.

• Pattern: each y-coordinate is the total a+b+c minus the matching x-coordinate.
• Consequence: so all three points lie on the single straight line x+y=a+b+c, and points on one line bound no area.
• Insight: recognising that line explains why the determinant collapses term by term to 0, rather than trusting a brute expansion.

Option (B), 0.

A square root opens two doors.

• Setup: squaring the distance gives 9+p²=25, so p²=16.
• Both roots: because squaring loses the sign, both p=4 and p=-4 sit at vertical distance 4 from (1,0) and give the same overall distance 5.
• Geometry agrees: one point lies above and one below, each equally far, so the honest answer keeps the ± sign.

Option (B), ± 4.

Slope from the origin is cleanest.

• Slope: since O is the origin, the line OA has slope 21=2.
• Condition: so C(a,b) lies on it only when ba=2, that is b=2a, and the area determinant gives the same 2a-b=0.
• Why simpler: anchoring on the origin turns the whole collinearity test into a single slope comparison.

Option (C), 2a=b.

DEF is ABC doubled.

• Scaling: each vertex of the second triangle is the first one pushed out from the origin by a factor of two, for instance A(-2,0) becomes D(-4,0).
• Why similar: a uniform scaling from a single point always produces a similar figure, so every side ratio equals two and the similarity is immediate.
• Role of the surds: the distance computations only confirm this enlargement, so you do not really need them once you spot the clean doubling of every vertex.

True; the scale factor is 2 throughout.

One coordinate test is enough here.

• Same line: the shared first coordinate pins all three points to one vertical line, so they are automatically collinear and no area or slope work is needed.
• Between the ends: the only thing left to decide is whether P lies between the two endpoints, and its height of two sits comfortably inside the range from the lower end up to the upper end.
• Verdict: since P is on the line and inside that range, it is a genuine interior point of the segment, which makes the statement true.

True; P is on the vertical line x=-4 between A and B.

Two on an axis, one off it.

• Their line: the first two points both live on the vertical axis, so the only straight line that can pass through both of them is that axis itself.
• The stray point: the third point sits three units to the right of the axis, so it misses that line completely, and the three of them simply cannot be collinear.
• Number check: the area determinant backs up the picture with a clear value of twenty-one square units, which is far from the zero that collinear points would need.

False; (3,6) is off the y-axis that holds the other two.

Compare the two distances directly.

• Definition: the perpendicular bisector is exactly the set of points that are equally far from A and from B.
• The numbers: from the given point the two distances come out as 2 and √10, which are plainly unequal, so the point does not lie on the bisector.
• Shortcut: there is no need to derive the bisector's equation, since a single pair of distance checks already settles the claim as false.

False; PA≠ PB.

The area collapses to zero.

• Determinant: plugging the three points into the area formula leaves -12+12=0 inside the bars, so the enclosed area vanishes.
• Geometry: the points therefore share one straight line, which leaves no region at all to call a triangle.
• Slope check: the step A→ B has slope -13 and so does A→ C, confirming that the three points line up.

True; the three points are collinear.

One midpoint pair decides it.

• Rule: for a parallelogram the two diagonals must meet at a single shared centre, so their midpoints have to agree.
• Here: the centre of one diagonal is (92,-32) while the centre of the other is (32,92), two entirely different points.
• Verdict: because those centres disagree the figure cannot be a parallelogram, and no further side computation is needed to rule it out.

False; the diagonals have different midpoints.

Radius first, then the point.

• Radius: the point on the circle fixes the radius at five units straight away.
• The query point: the point being tested sits a clean ten units from the origin, another 6-8-10 triple, which is double the radius.
• Verdict: being twice as far out as the boundary puts it firmly outside the circle, so the statement holds true.

True; OQ=10 is greater than the radius 5.

Both distances are multiples of 5.

• Compute: the two distances come out as 25 and 55, both built on the same surd.
• Mismatch: sharing that common unit makes the gap obvious at a glance, two against five, so the distances are clearly unequal.
• Verdict: unequal distances put the point off the perpendicular bisector, so the claim is false, and the common surd unit again exposes the comparison cleanly.

False; AP≠ AQ.

The 1:2 split lands on P.

• Meaning: trisection cuts the segment into three equal parts, so the two dividing ratios are 1:2 and 2:1.
• Test: feeding the 1:2 ratio into the section formula reproduces the point (5,-3) exactly, which is the given P.
• Verdict: since P matches a trisection point the claim is true, and the other trisection point would come from the 2:1 ratio instead.

True; P divides AB as 1:2.

Everything sits in 5 units.

• Three lengths: the distances are 25, 75 and 95, all multiples of the same surd.
• Between: their parts add as two plus seven equals nine, which shows B lies between A and C, so the points are collinear.
• The ratio: the shorter to the whole is two to nine, giving AB=29AC directly, so both halves of the statement check out and it is true.

True; AB=29AC and AB+BC=AC.

Square both to compare cleanly.

• The squares: the squared distance from the centre is 25+1=26, while the squared radius is 36.
• Comparison: since twenty-six is less than thirty-six, the point is closer to the centre than the boundary, so it lies inside rather than on the circle.
• Why squares: comparing the squared values avoids approximating the awkward √26 and still gives the verdict at a glance.

False; CP=√26, less than the radius 6.

Two conditions, both met.

• Shared centre: the diagonals share the midpoint (12,32), which already makes the figure a parallelogram.
• Equal length: they are also of equal length √58, and a parallelogram with equal diagonals must have right angles, so it upgrades to a rectangle.
• Why this route: checking one midpoint and the two diagonal lengths is the quickest coordinate path to confirming a rectangle.

True; equal, mutually bisecting diagonals.

Compare squared lengths.

• Skip the surds: working with the squared side lengths 65, 170 and 145 avoids square roots entirely.
• All distinct: those three values are clearly different, and distinct sides mean a scalene triangle.
• Bonus check: a right-angle test is worth a glance too, since 65+145=210170, so the triangle is not right angled, just plainly scalene.

Scalene triangle.

Set y=0, then solve the square.

• Fix the height: any point on the x-axis is (x,0), so the vertical gap to (7,-4) is locked at four.
• Solve: squaring the distance leaves (x-7)²=20-16=4, which has the two roots x=9 and x=5.
• Geometry: a circle of the given radius about (7,-4) cuts the x-axis in two places, which matches the two solutions you found.

(9,0) and (5,0); two points.

Two pairs of equal sides, two equal diagonals.

• Opposite sides: the matching pairs √50 and √32 tell you straight away that the figure is at least a parallelogram.
• Diagonals: both come out equal at √82, and equal diagonals in a parallelogram force the corners to be right angles, which lifts it to a rectangle.
• Not a square: the adjacent sides differ in length, so the shape is a rectangle and never a square, which would need all four sides the same.
• Order matters: keep the vertices in the listed order A,B,C,D so that you pair genuine sides with sides and genuine diagonals with diagonals, not a mix of the two.

Rectangle.

The vertical gap is already the full distance.

• Spot it: the difference between the two heights is exactly nine, which is the whole distance the question demands.
• Consequence: if the vertical part alone uses up the entire length, then the horizontal part must contribute nothing at all.
• Solve: that zero horizontal gap means a+3=0, so a=-3, with no algebra beyond reading the coordinates.
• Single answer: because the points end up stacked on the line x=-3, only one value of a works, unlike the usual two-root distance problems.

a=-3.

Name one point, then count the whole line.

• Easy sample: the midpoint of the segment is the quickest equidistant point to write down, and here it works out to (-3,5).
• Verify: a quick distance check confirms it, since the gap to each given point is the same value 5, so the midpoint really does qualify.
• The full set: being the same distance from two points actually defines an entire straight line, the perpendicular bisector of the segment, not just a single spot.
• Two-part answer: so the honest reply names one concrete sample point and then states that infinitely many such points lie along that bisector.

(-3,5); infinitely many such points.

A flat segment has an upright bisector.

• Why vertical: the two points sit at the same height, so the segment joining them is horizontal and its perpendicular bisector must stand straight up.
• Its position: that upright line passes through the midpoint, so it is the line x=-12, fixed by the shared midpoint of the two points.
• Find Q: where this line meets the horizontal axis the height is zero, which pins the required point at (-12,0).
• Triangle type: any point on a bisector is equally far from the two ends, so the new point gives a triangle with two equal sides, namely an isosceles one.

Q=(-12,0); isosceles triangle.

Zero area gives one clean linear equation.

• Why zero: collinear points enclose no triangle, so the area bracket must vanish, which turns the area formula into an equation you can solve.
• Solve: substituting the points and tidying leaves 19-14m=0, a single linear equation that gives m=1914 at once.
• Slope check: the line through the two fixed points has slope 47, and forcing the third point to share that slope reproduces the very same value of m.
• Takeaway: either route, area or slope, works, but the area method is usually quicker because it avoids fractions until the final step.

m=1914.

Square the distances, then split into cases.

• Cancel: equating the two squared distances wipes out the common term and leaves the tidy equation (y+4)²=1, free of any surds.
• Two roots: that square equation has two solutions, y=-3 and y=-5, each placing the point Q at a different spot on its line.
• Two distances: because Q moves, the length PQ has to be worked out twice, giving √290 in one case and 132 in the other.
• Report both: the question allows two valid positions, so dropping either case would lose half the correct answer.

y=-3 or -5; PQ=√290 or 132.

The three points secretly share one line.

• Cancel to zero: the area determinant collapses to 24-30+6=0, which is the clear signal that the points are collinear.
• See the pattern: every step of two to the right raises the height by two, so the slope is one and all three points lie on the line y=x+12.
• No region: points strung along one straight line enclose no space, so there is genuinely no triangle to measure here.
• Honest answer: the only correct value is zero square units, and the lined-up pattern explains why, rather than leaving it as a bare arithmetic accident.

Area =0; the points are collinear.

One fact drives the whole problem: the axis has zero height.

• Key idea: any point on the horizontal axis has height zero, so set the section formula's height part equal to zero from the start.
• Find the ratio: that condition reduces to 7k-6=0, giving k=67, which is the ratio 6:7 in which the axis cuts the segment.
• Find the point: feeding that same ratio into the width part of the formula gives the crossing point at x=-3413.
• Why it is quick: you never need the full division for both coordinates, since the zero height already settles the ratio in one line.

6:7, at (-3413,0).

Solve with one coordinate, confirm with the other.

• Set up: write the unknown ratio as k:1 so the section formula carries just one letter, keeping the algebra short.
• Solve: matching the width coordinate gives 5k=1, so k=15, which scales up to the whole-number ratio 1:5.
• Confirm: the height coordinate then lands on exactly 512, proving the point sits on the segment itself and not on the line stretched beyond it.
• Worth the check: with fractional coordinates a slip is easy, so the matching second coordinate is reassurance well worth the extra line.

1:5.

The two unknowns separate cleanly, one per coordinate.

• Width gives a: the width equation contains only the letter a and tidies to 9a=9, so a=1 with no other unknown getting in the way.
• Height gives b: the height equation contains only b and reduces to -b=3, which hands you b=-3 straight away.
• No simultaneous work: because the unknowns never mix, you never have to solve two equations together, just two short linear steps in turn.
• General habit: whenever a single point equation appears, split it into its width and height parts first and check whether the unknowns fall apart like this.

a=1, b=-3.

Pin the easy value first, then unwind to the rest.

• Free value: the height of the midpoint depends only on known numbers, so it is fixed at -1 no matter what k turns out to be.
• Use the constraint: feeding that height into the given condition a-2b=18 immediately gives a=16, with nothing else to solve.
• Recover k: putting a=16 back into the width half of the midpoint gives k=22, and the distance formula then returns AB=2√61.
• Why this order: working b, then a, then k, then the distance keeps every stage a single tidy line instead of a tangled system.

k=22, AB=2√61.

The centre-to-point distance must equal the radius.

• Radius first: halve the diameter to get a radius of 52, whose square is the tidy number 50 that the equation will use.
• Set up: demand that the moving centre stay this fixed distance from the given point, which after expanding becomes the quadratic a²-8a+15=0.
• Factor: it splits as (a-3)(a-5)=0, so two values of the unknown both satisfy the distance condition.
• Keep both: each value describes a different but valid centre sitting the correct distance from the point, so neither root should be thrown away.

a=3 or a=5.

Divide first, then substitute into the line.

• Find the point: the section formula with the given ratio places the division point P at (113,53).
• Substitute: drop those coordinates into the line equation, which is now just a plug-in with no fresh geometry needed.
• Clean cancels: the fractions collapse neatly, since 3·113=11 and 18·53=30, leaving the short equation 11-30+k=0, so k=19.
• Self-check: those tidy cancellations are a good sign the ratio was applied the right way round, since a misapplied ratio usually leaves ugly fractions.

k=19.

Skip hunting for the three original vertices.

• The shortcut: the triangle formed by the three midpoints is similar to the original one, with every side exactly half as long.
• Area ratio: halving each side scales the area by a factor of one quarter, so the midpoint triangle holds a quarter of the area you want.
• Apply it: compute the small triangle's area as 114 and multiply by four, which gives the answer of eleven square units in a single step.
• Why it saves work: the alternative of solving for the three corners first is far longer, so spotting the quarter-area rule is the real labour-saver here.

11 square units.

A right angle hands you a Pythagoras equation.

• Set up: the right angle at B means the two short sides squared add to the longest side squared, which gives the quadratic a²-7a+10=0.
• Roots: this factors to give a=2 or a=5, so two values of the unknown solve the equation as it stands.
• Reject one: the value a=5 makes two corners land on the same point and flatten the triangle, so only a=2 leaves a real figure.
• Instant area: at the right angle the two legs are the perpendicular sides 3 and 4, so the area is just half their product, namely six square units.

a=2; area =6 square units.

Turn the fraction of a segment into a clean ratio.

• Read it right: being three-fifths of the way along from the first point means the point splits the segment as three parts to two, not three to five.
• Apply the ratio: the section formula with that three to two split gives the point at (45,215) directly.
• The common slip: students often write the ratio as three to five, but the second number must be the leftover part of the segment, which is two.
• Safe rule: a fraction such as three-fifths always becomes the ratio numerator to the difference, here three to two, before you touch the formula.

R=(45,215).

Zero area turns into a quadratic in the unknown.

• Set up: since the points must be collinear, the area is zero, and substituting the variable vertices gives the quadratic 6k²-15k+6=0.
• Simplify: divide through by the common factor to get the tidier 2k²-5k+2=0, which is easier to factor.
• Solve: it factors as (2k-1)(k-2)=0, so the two solutions are k=12 and k=2.
• Keep both: each value slides all three points to new spots but still leaves them on one straight line, so both answers are genuine.

k=12 or k=2.

Force the section point to obey the line.

• Set up: write the dividing point in terms of the unknown ratio, then demand that it satisfy the given line equation.
• Solve: that demand clears all the fractions and leaves the simple linear equation 2k-16=0, so k=8 and the ratio is eight to one.
• Find the point: feeding that value back into the formula gives the division point at (83,-19).
• Internal cut: because the ratio came out positive, the line crosses inside the segment, which the point lying between the two endpoints confirms.

8:1, at (83,-19).

Symmetry fixes the width, side equality fixes the height.

• Use symmetry: the base is balanced about the vertical axis, so the apex of an equilateral triangle has to sit on that axis, with width coordinate zero.
• Side equality: setting a slant side equal to the base length of eight gives the equation (y-3)²=48, and so two heights y=343 both fit.
• Two apexes: one solution lifts the apex above the base and the other drops it below, so the bare algebra cannot decide between them on its own.
• Use the clue: the base sits at height three, and only the lower apex near 3-43, which is about negative four, wraps the triangle around the origin, so the interior condition picks the negative root.

(0, 3-43).

Build one point at a time, in a fixed order.

• Find the corner: the diagonals of a parallelogram bisect each other, which gives the missing vertex as D=A+C-B=(7,3) in a single line.
• Find the midpoint: averaging that corner with C places the midpoint at (8,72), the next point the question needs.
• Find the area: feeding the three known points into the area determinant returns 34 of a square unit.
• Why it is easy: each stage uses just one standard tool, so a question that looks long really breaks into three short and separate computations.

34 square units.

One formula, run three times on three medians.

• Where a median lands: the foot of each median is just the midpoint of the opposite side, so finding it costs only one average.
• The two to one point: dividing a median two to one from its vertex always simplifies down to the plain average of all three corner coordinates.
• Same answer thrice: doing that section algebra on each of the three medians returns the very same point (x₁+x₂+x₃3,y₁+y₂+y₃3) every time.
• What it proves: that the three points coincide is exactly the theorem that the medians meet at one place, the centroid, which cuts each median in the ratio two to one.

Centroid =(x₁+x₂+x₃3,y₁+y₂+y₃3), and P=Q=R equal it.

Find the unknown first, then peel off the height.

• Fix a: matching the midpoints of the two diagonals, which must coincide in a parallelogram, gives a=-3 straight away.
• Get the area: the whole parallelogram has twice the area of the triangle on three of its corners, which comes to 24 square units.
• Get the height: dividing that area by the base length √26 and rationalising leaves a height of 12√2613.
• Why this order: vertex first, then area, then height keeps each computation self-contained, so a mistake in one stage never spreads into the next.

a=-3; height =12√2613.

The diagonal crossing is the spot you want.

• Read the grid: the four students sit at the corners of a square, with the columns giving the width values and the rows giving the height values of each position.
• What the spot must be: a point the same distance from all four corners is the centre of the circle through them, and for a square that centre is simply where the two diagonals cross.
• Cross the diagonals: one diagonal runs flat and the other runs upright, and both have the same midpoint, so they meet at the single point (7,5), confirming the figure is a true square.
• Measure out: from that centre the corners lie four units to the left, right, above and below, so every student is the same distance away and the matching four lengths prove the point is genuinely equidistant rather than merely close.

Yes; position (7,5), which is 4 units from each student.

Add three neat legs, then subtract the shortcut.

• Tidy legs: each stretch of the walk is a whole number, namely five and ten from familiar right-triangle triples and twelve from a straight vertical hop.
• Detour total: adding the three legs gives twenty-seven kilometres for the route that visits the bank and the school on the way.
• Direct route: the straight hop from home to office is shorter at √605=115, which is about twenty-four and a half kilometres.
• Extra walking: the gap between the two routes, roughly two and a half kilometres, is the extra distance, and spotting the right-triangle triples makes every leg instant.

About 2.4 km extra.