Question

A straight line through Q(2,3) makes an angle with 3 180 ÷ 4 with the negative direction of x- axis if the straight line intersects the line x+y-7 is equal to zero at p find distance pq

Answer 1

Answer:

The distance PQ = $\boldsymbol{\sqrt{2}}$

Step-by-step explanation:

Since the line passing through the point Q (2, 3) makes and angle with the negative x-axis = $3\times\frac{180^\circ}{4} =135^\circ$

Therefore with positive x-axis it will make an angle = 180° - 135° = 45°

Therefore, the slope of the line m = tan45° = 1

Let the equation of the line be

y = mx + c

Here m = 1

hence, y = x + c

Since this line passes through point (2, 3), therefore (2, 3) will satisfy the above equation

Hence

3 = 2 + c

or c = 1

Hence the equation of the line becomes y = x + 1

or, x - y = -1  ......... (1)

Again the equation of the other straight line given

x + y - 7 = 0

or, x + y = 7   .......... (2)

In order to find the point of intersection P of both the lines, we need to solve the equation (1) and (2)

Adding the equation (1) and (2)

2x = 6

⇒ x = 3

Putting the value of x in eq (2)

3 + y = 7

or, y = 7 - 3 = 4

Therefore the coordinates of point P is (3, 4)

Now we know that if two points are (x₁, y₁) and (x₂, y₂) then the distance between the two points is given by

$\boxed{d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} }$

Therefore, the distance

$PQ=\sqrt{(3-2)^2+(4-3)^2} =\sqrt{1+1} =\sqrt{2}$

Hope this helps.