Question

5. 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:

hope u know the formulea so i didnt write

Answer 2

$2\sqrt{2}$ is the distance PQ.

Explanation:

Given that,

A straight has slope(m) = 1

Q = (-3, 5)

As we know,

The equation of line:

$(x - x) = m(y - y)$

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

∵ $x - y + 8 = 0$

A.T.Q.

The line $(x - y + 8 = 0)$ at Q meets $(x + y -6 = 0)$ at P

by solving both of them, we get

M   L     F      M

-1    8     1      -1

1    -6     1      1

____________

$x/(6 - 8) = y/(8 + 6) = 1/(1 + 1)$

⇒ $x/-2 =y/14 = 1/2$

∵$x = -1, y = 7$

Thus, P = (-1, 7)

∵ PQ = $\sqrt{(2)^2 + (2)^2}$

= $2\sqrt{2}$

Learn more: Find the distance

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