Find the equation of a straight line.Passing through the point of inte

1.	Find the equation of a straight line.Passing through the point of intersection of two lines 3x – 5y = 1 and 2x + 3y = 7 and the point (-4, 3).
Answer» $\large\underline{\bold{Solution-}}$ Any line which passes through the point of intersection of the lines 3x – 5y - 1 = 0 and 2x + 3y - 7 = 0 is given by $\sf \: 3x - 5y - 1 + k(2x + 3y - 7) = 0$ $\rm :\longmapsto\:3x - 5y - 1 + 2kx + 3ky - 7k = 0$ $\rm :\longmapsto\:(3 + 2k)x + ( - 5 + 3k)y - (7k + 1) = 0 - - (1)$ Now, equation (1) passes through (- 4, 3), then $\rm :\longmapsto\:(3 + 2k)( - 4)+ ( - 5 + 3k)(3) - (7k + 1) = 0$ $\rm :\longmapsto\: - 12 - 8k - 15 + 9k - 7k - 1 = 0$ $\rm :\longmapsto\: - 6k - 28 = 0$ $\rm :\longmapsto\: - 6k = 28$ $\bf\implies \:k \: = \: - \: \dfrac{14}{3}$ Now, Substitute the value of 'k' in equation (1), we get $\sf \: \bigg( 3 + 2 \times \dfrac{( - 14)}{3} \bigg)x + \bigg( - 5 + 3 \times \dfrac{( - 14)}{3} \bigg) y- \bigg( 7 \times \dfrac{( - 14)}{3} + 1 \bigg) = 0$ $\sf \: \bigg( \dfrac{9 - 28}{3} \bigg)x + \bigg( \dfrac{ - 15 - 42}{3} \bigg)y - \bigg( \dfrac{ - 98 + 3}{3} \bigg) = 0$ $\sf \: - 19x - 57y + 95 = 0$ Divide whole equation by (- 19), we get $\bf :\longmapsto\:x + 3y - 5 = 0$ Thus, The required equation of line which passes through the point of intersection of the lines 3x – 5y = 1 and 2x + 3y = 7 and the point (- 4, 3) is x + 3y - 5 = 0. 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, the equation of line is given by x cosβ + y sinβ = p.

Find the equation of a straight line.Passing through the point of intersection of two lines 3x – 5y = 1 and 2x + 3y = 7 and the point (-4, 3).

Answer»

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

Any line which passes through the point of intersection of the lines 3x – 5y - 1 = 0 and 2x + 3y - 7 = 0 is given by

$\sf \: 3x - 5y - 1 + k(2x + 3y - 7) = 0$

$\rm :\longmapsto\:3x - 5y - 1 + 2kx + 3ky - 7k = 0$

$\rm :\longmapsto\:(3 + 2k)x + ( - 5 + 3k)y - (7k + 1) = 0 - - (1)$

Now, equation (1) passes through (- 4, 3), then

$\rm :\longmapsto\:(3 + 2k)( - 4)+ ( - 5 + 3k)(3) - (7k + 1) = 0$

$\rm :\longmapsto\: - 12 - 8k - 15 + 9k - 7k - 1 = 0$

$\rm :\longmapsto\: - 6k - 28 = 0$

$\rm :\longmapsto\: - 6k = 28$

$\bf\implies \:k \: = \: - \: \dfrac{14}{3}$

Now, Substitute the value of 'k' in equation (1), we get

$\sf \: \bigg( 3 + 2 \times \dfrac{( - 14)}{3} \bigg)x + \bigg( - 5 + 3 \times \dfrac{( - 14)}{3} \bigg) y- \bigg( 7 \times \dfrac{( - 14)}{3} + 1 \bigg) = 0$

$\sf \: \bigg( \dfrac{9 - 28}{3} \bigg)x + \bigg( \dfrac{ - 15 - 42}{3} \bigg)y - \bigg( \dfrac{ - 98 + 3}{3} \bigg) = 0$

$\sf \: - 19x - 57y + 95 = 0$

Divide whole equation by (- 19), we get

$\bf :\longmapsto\:x + 3y - 5 = 0$

Thus,

The required equation of line which passes through the point of intersection of the lines 3x – 5y = 1 and 2x + 3y = 7 and the point (- 4, 3) is x + 3y - 5 = 0.

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