Question

A straight line with slope 1 passes through Q(-3,5) and meets the straight line x+y-6 = 0 at P. find the distance PQ​

Answer 1

Answer:

$2 \sqrt{2}$

Step-by-step explanation:

y - y1 = m(x - x1)  is the formula for line passing through a point

y -5 = 1(x - (-3))

y-5 =x +3

y =  x + 8

to find y= x+8 and x + y = 6  equate both equations

y = x + 8

x + y = 6

x +(x+8) = 6

2x +8 = 6   divide both sides by 2

x+4 = 3

x = -1

y = -1 +8 = 7

point is P(-1,7)

distance between P(-1,7)  and Q(-3 ,5)

distance = $\sqrt{(x2 -x1)^{2} + (y2 -y1)^{2} }$

$=\sqrt{(-3 - (-1))^{2} + (5-7)^{2} } \\=\sqrt{(-2)^{2} + (-2)^{2} }\\=\sqrt{8} =2\sqrt{2}$

Answer 2

Answer:

Step-by-step explanation: