Tag: Class 10th Exercise 7.1 Maths NCERT Solution

1. Find the distance between the following pairs of points :

(i) (2, 3), (4, 1)
Distance = $\sqrt{(4-2{)}^2+(1-3{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\mathrm{.}$

(ii) (−5, 7), (−1, 3)
Distance = $\sqrt{(-1+5{)}^2+(3-7{)}^2}=\sqrt{4^2+(-4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\mathrm{.}$

(iii) (a, b), (−a, −b)
Distance = $\sqrt{(-a-a{)}^2+(-b-b{)}^2}=\sqrt{(-2a{)}^2+(-2b{)}^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}\mathrm{.}$

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2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Distance = $\sqrt{{36}^2+{15}^2}=\sqrt{1296+225}=\sqrt{1521}=39.$

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3. Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.

Compute slopes:
slope between (1,5) and (2,3) = $(3-5)\mathrm{/}(2-1)=-2.$
slope between (1,5) and (−2,−11) = $(-11-5)\mathrm{/}(-2-1)=-16\mathrm{/}-3=16\mathrm{/}3.$

Slopes are not equal, so the three points are not collinear.

(You can also check by area of triangle ≠ 0.)

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4. Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.

Compute side-lengths:

• $AB=$distance between (5,−2) and (6,4) = $\sqrt{1^2+6^2}=\sqrt{37}\mathrm{.}$

• $BC=$distance between (6,4) and (7,−2) = $\sqrt{1^2+(-6{)}^2}=\sqrt{37}\mathrm{.}$

• $AC=$distance between (5,−2) and (7,−2) = $\sqrt{2^2+0^2}=2.$

Two sides $AB$ and $BC$ are equal ($\sqrt{37}$), so yes — they form an isosceles triangle (vertex at $B$).

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5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli … “Don’t you think ABCD is a square?” … Using distance formula, find which of them is correct.

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6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) (−1, −2), (1, 0), (−1, 2), (−3, 0)

Compute side lengths and diagonals (worked out):
All four sides ≈ $2\sqrt{2}$ and both diagonals = 4. Opposite sides equal and all four sides equal, diagonals equal ⇒ this quadrilateral is a square.

(ii) (−3, 5), (3, 1), (0, 3), (−1, −4)

Checking points in the given order shows that A, B, C are collinear (area of triangle ABC = 0). So the four given points do not form a proper (non-degenerate) quadrilateral — the three first points lie on a line (degenerate case). Hence no quadrilateral (degenerate).

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Slopes (consecutive): AB and CD have equal slope (both $1\mathrm{/}3$), BC and DA have equal slope (both $1$). Opposite sides are parallel and opposite sides equal in length ⇒ a parallelogram (not a rectangle or rhombus — adjacent sides unequal).

(If you want, I can show the numeric distances/ slopes step-by-step for each case.)

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7. Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).

Let the point be $(x,0)$. Equate squared distances:

$(x-2{)}^2+(0+5{)}^2=(x+2{)}^2+(0-9{)}^2$
$(x-2{)}^2+25=(x+2{)}^2+81$ → solve → $x=-7.$

So the required point is $(-7,0)$.

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8. Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.

$\sqrt{(10-2{)}^2+(y+3{)}^2}=10$ → $64+(y+3{)}^2=100$ → $(y+3{)}^2=36$ → $y+3=\pm 6$.

Therefore $y=3$ or $y=-9.$

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9. If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Equate squared distances:

$(0-5{)}^2+(1+3{)}^2=(0-x{)}^2+(1-6{)}^2$
Left: $25+16=41$. Right: $x^2+25$. So $x^2+25=41$ → $x^2=16$ → $x=4$ or $x=-4$.

• For x = 4:
  $QR=\sqrt{(0-4{)}^2+(1-6{)}^2}=\sqrt{16+25}=\sqrt{41}\mathrm{.}$
  $PR=$distance between P(5,−3) and R(4,6) $=\sqrt{1^2+9^2}=\sqrt{82}\mathrm{.}$

• For x = −4:
  $QR=\sqrt{(0+4{)}^2+(1-6{)}^2}=\sqrt{41}\mathrm{.}$
  $PR=$distance between P(5,−3) and R(−4,6) $=\sqrt{9^2+9^2}=\sqrt{162}=9\sqrt{2}\mathrm{.<span\; class="mord\; sqrt"\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span></span></span>}$

So $Q$ is equidistant to $P$ and $R$ for $x=\pm 4$; $QR=\sqrt{41}$ in both cases; $PR$ differs as above.

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10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).

Equate squared distances:

$(x-3{)}^2+(y-6{)}^2=(x+3{)}^2+(y-4{)}^2$

Expand and simplify ⇒ $-12x-4y+20=0$ ⇒ dividing by −4 ⇒ $3x+y-5=0$.

So the required relation is $3x+y-5=0$ (or $y=5-3x$.