Question

Find the distance of the point (2,5) from the line 3x + y + 4 = 0 measured parallel to the line having slope 3/4 ?

Need Step by step explanation

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The distance of the point (2,5) from the line 3x + y + 4 = 0 is 5 units.

Step-by-step explanation:

Given:

As shown in the figure, let P (2,5) = ($x_1,y_1$) be the point from which we are measuring the distance .

The equation of the line passing through the point P and having slope $\frac{3}{4}$ is,

$y-y_1=m(x-x_1)$

$y-5=\frac{3}{4} (x-2)$

4y-20=3x-6

3x-4y+14=0                   (1)

Now, at point A, let this line intersect the given line 3x+y+4=0     (2)

therefore the required distance will be AP.

Subtracting equation (2) from (1)

3x-4y+14-(3x+y+4)=0

3x-4y+14-3x-y-4=0

-5y+10=0

5y=10

$y=\frac{10}{5}$

y=2

substitute the value of y in equation (2)

3x+y+4=0

3x+2+4=0

3x+6=0

3x=-6

$x=-\frac{6}{3}$  

x=-2

Therefore the co-ordinates of the point A is (-2,2). and we have P (2,5).

Distance,

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

$AP=\sqrt{(-2-2)^2+(2-5)^2}$

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

$AP=\sqrt{16+9}$

$AP=\sqrt{25}$

AP=5

Hence AP= 5 units.