Question

Q2. Find the equation of the line passing through the point of intersection of the lines 4x + 7y - 3 = 0 and 2x - 3y + 1 = 0 that has equal intercepts on the axes.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that line passes through the the point of intersection of the lines 4x + 7y - 3 = 0 and 2x - 3y + 1 = 0.

The required equation of line is given by

$\rm :\longmapsto\:4x + 7y - 3 + k(2x - 3y + 1) = 0$

$\rm :\longmapsto\:4x + 7y - 3 + 2kx - 3ky + k = 0$

$\rm :\longmapsto\:(4 + 2k)x + (7 - 3k)y + k - 3 = 0$

$\rm :\longmapsto\:(4 + 2k)x + (7 - 3k)y = 3 - k - - (1)$

Now, we reduce this equation to intercept form

Divide both sides by 3 - k, we get

$\rm :\longmapsto\:\dfrac{4 + 2k}{3 - k} \: x + \dfrac{7 - 3k}{3 - k} \: y = 1$

$\rm :\longmapsto\:\dfrac{x}{ \: \: \: \dfrac{3 - k}{4 + 2k} \: \: \: } + \dfrac{y}{ \: \: \: \dfrac{3 - k}{7 - 3k} \: \: \: } = 1$

which is the required intercept form of the line.

Now,

Intercept made by the line is given by

$\rm :\longmapsto\:Intercept_{(x - axis)} = \dfrac{3 - k}{4 + 2k}$

and

$\rm :\longmapsto\:Intercept_{(y - axis)} = \dfrac{3 - k}{7 - 3k}$

According to statement,

$\rm :\longmapsto\:Intercept_{(x - axis)} = Intercept_{(y - axis)}$

$\rm :\longmapsto\:\dfrac{3 - k}{4 + 2k} = \dfrac{3 - k}{7 - 3k}$

$\rm :\longmapsto\:\dfrac{1}{4 + 2k} = \dfrac{1}{7 - 3k}$

$\rm :\longmapsto\:4 + 2k = 7 - 3k$

$\rm :\longmapsto\: 2k + 3k = 7 - 4$

$\rm :\longmapsto\: 5k = 3$

$\rm :\longmapsto\:k = \dfrac{3}{5}$

On substituting the value of k, in equation (1), we get

$\rm :\longmapsto\:(4 + 2 \times \dfrac{3}{5} )x + (7 - 3 \times \dfrac{3}{5} )y = 3 - \dfrac{3}{5}$

$\rm :\longmapsto\:(4 + \dfrac{6}{5} )x + (7 - \dfrac{9}{5} )y = \dfrac{15 - 3}{5}$

$\rm :\longmapsto\:( \dfrac{20 + 6}{5} )x + ( \dfrac{35 - 9}{5} )y = \dfrac{15 - 3}{5}$

$\rm :\longmapsto\:26x + 26y = 12$

$\bf :\longmapsto\:13x + 13y = 6$

Additional Information

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line

2. Point-slope form equation of line

Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by

y - b = m(x - a)

3. Slope-intercept form equation of line

Consider a line whose slope is m which cuts the Y-axis at a distance ‘a’ from the origin. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by

y = mx + a.

4. Intercept Form of Line

Consider a line L having x– intercept a and y– intercept b, then the line passes through  X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1.

5. Normal form of Line

Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis.

Then, equation of line is given by x cosβ + y sinβ = p.

Answer 2

How do you find the equation of the line passing through the point of intersection of the lines 4x +7y -3 = 0 and 2x -3y +1 = 0 that has equal intercepts on the axes?

Solving these two lines for x and y

Multiplying EQ of 2nd line by 2 then subtracting from 1st line

4x+7y-3–4x+6y-2=0

13y=5

y=5/13

Putting the value of y in 2nd equation

2x-15/13+1=0

2x=15/13–1=2/13

x=1/13,so third line passes through (1/13,5/13)

If the required line makes equal intercept on the axis then slope will be either 1 or —1

Eq of required line y-5/13=1(x-1/13)I.e.13x-13y+4=0

And y-5/13=—1(x-1/13)

13x+13y-6=0