Question

9. Find the equation of the lines passing through the point (4, 0) and equally inclined to the co-ordinate axes. Sol. (i) Poil+S-32 YA Bonil a te coltepa sdt boi to L 45 # A(4,0) X S ==​

Answer 1

Answer:

First of all you did not mention the sign in the equation 3x=4y_7, so now I have to take two cases.

Case 1:

3x=4y+7 and 5y=12x+6

Now we are suppose to find the equation of a line passing via coordinate (4,5) and it is equally inclined to both the above mentioned lines. Therefore, we can say that the required line is the angle bisector of the above mentioned lines.

Now let us find the intersection point of the two lines:

3x=4y+7 and 5y=12x+6

By doing cross multiplying method we get x=(-54/33) and y=(-34/11). Thus the point of intersection is (-54/33, -34/11).

Now, the equation of line which we intend to find must pass through (4,5) and (-54/33, -34/11) to be the angle bisector.

Hence we find the slope (m) of the line:

m={[5+(34/11)]/[4+(54/33)]}

m=267/191

Now we know that y=mx+c and we have a point and slope, so by replacing those values we can find c:

5=(267/191)4 +c

c= -113/191

Therefore, the equation of the line is :

y=(267/191)x - (113/191)

191y=267x-113

Case 2:

3x=4y-7 and 5y=12x+6

Now if we put the point (4,5) in 3x=4y-7 we get:

3(4) = 4(5) -7

12 = 20–7

12 = 13

Answer 2

Step-by-step explanation:

4,0 is the correct answer

hope it helps you