NCERT Exemplar Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 Solutions

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.