Answer࿐

Given,

3X+2y = -4 ----------(1)

2x+5y = 1 -----------(2)

¶ FIND POINT of Intersection of these 2 lines

Do 2×(1) - 3×(2)

6x+4y = -8 ----------(3)

6x+15y = 3 ---------(4)

-----------------------

-11y = -11

=> y = 1

Substitute in (1)

3x+2(1) = -4

=> 3x = -4-2

=> 3x = -6

=> X = -2

•°• Point of Intersection of the lines 3x+2y+4= 0 & 2x+5y-1= 0 is (-2,1)

¶ By using the Slope-Intercept form, find the form of equation of line passing through the point (-2,1)

y = mx + c

substitute x = -2 & y = 1

=> 1 = -2m + c

=> c = 1 + 2m

•°• The Required Equations of straight line is of form :

y = mx + 1 + 2m -----------(5)

¶ The perpendicular distance (or simply distance) 'd' of a point P(x1,y1) from Ax+By+C = 0 is given by

Given,

(x1,y1) = (-2,1) & d = 2

=> (4m+2)² = 2(m² + 1)

=> 16m² + 16m + 4 = 2m² + 2

=> 14m² + 16m + 2 = 0

=> 7m² + 8m + 1 = 0

Factorise the equation

=> 7m² + 7M + m + 1 = 0

=> 7m(m+1) + 1(m+1) = 0

=> (m+1)(7m+1) = 0

=> m = -1 and m = -1/7

Now Substitute m = -1 in (5)

=> y = (-1)x+1+2(-1)

=> y = -x + 1 - 2

=> y = -1 - x

(or)

=> -x - y -1 = 0

=> x + y + 1 = 0 ------------(6)

Substitute m = -1/7 in (5)

=> y = (-1/7)x + 1 + 2(-1/7)

=> y = -x/7 + (7-2)/7

=> y = (-x+5)/7

=> 7y = -x+5

or

=> -x - 7y + 5 = 0

=> x + 7y - 5 = 0 ----------(7)

•°• The Required equation of straight lines are :

x + y + 1 = 0 & x + 7y - 5 = 0

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