Question

P is a point whose ordinate and abscissa are same. Q ≡ (7, 11). If length of PQ = 20, find the coordinates of P.

Answer 1

Given:

P = (x, x) 

Q = (7, 11)

PQ = 20 

Step 1: Use Distance Formula

\[PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

Substitute P = (x, x), Q = (7, 11), and (PQ = 20)

\[ 20 = \sqrt{(7 - x)^2 + (11 - x)^2}\]

20² = (7 − x)² + (11 − x)²     ... [Square both sides]

400 = (7 − x)² + (11 − x)²

(7 − x)² = (x − 7)²       ... [Expand the squares]

= x² − 14x + 49

(11 − x)² = (x − 11)²

= x² − 22x + 121

400 = x² − 14x + 49 + x² − 22x + 121

400 = 2x² − 36x + 170

2x² − 36x + 170 = 400

2x² − 36x + 170 − 400 = 0

2x² − 36x − 230 = 0

x² − 18x − 115 = 0    .... [Divide entire equation by 2]

x² − 18x − 115 = 0

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

where (a = 1, b = −18, c = −115).

Δ = (−18)² − 4(1) (−115)

= 324 + 460 = 784

= \[\sqrt{784}\]

= 28

 \[x = \frac{18 \pm 28}{2}\]

\[x = \frac{18 + 28}{2}\]

= \[\frac{46}{2}\]

= 23

= \[x = \frac{18 - 28}{2}\]

= \[\frac{-10}{2}\]

= −5

The coordinates of (P) are P = (23, 23) or P = (−5, −5)