If the coordinates of the vertex A of a square ABCD are (3, -2).

Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test

⇒ y − y₁= m(x − x₁)

⇒ y−(−2)= − $\frac{7}{3}$ (x−3)

⇒ 3(y + 2) =−7(x−3)

⇒ 3y + 6 =−7x + 21

⇒ 7x + 3y = 15

Now we will find the coordinates of O, the points of intersection of AC and BD

⇒ 3x – 7y = -6 ….(i)
⇒ 7x + 3y = 15 ….(ii)

Multiplying (i) by 3 and (ii) by 7, we get,

⇒ 9x – 21y = -18 ….(iii)
⇒ 49x + 21y = 105 …(iv)

Adding (iii) and (iv) we get,

⇒ 9x + 49x – 21y + 21y = -18 + 105

⇒ 58x = 87

⇒ x = $\frac{87}{58}=\frac{3}{2}$ .

⇒ 3 × $\frac{3}{2}$ − 7y=−6

⇒ $\frac{9}{2}$ −7y=−6

⇒ $\frac{9}{2}$ + 6 = 7y

⇒ $\frac{9+12}{2}$ = 7y

⇒ $\frac{21}{2}$ = 7y

⇒ y = $\frac{3}{2}$

Hence, the equation of AC is 7x + 3y – 15 = 0 and coordinates of the center are $\left(\frac{3}{2}, \frac{3}{2}\right)$

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