# EQUATIONS OF LINES ADVANCED QUESTIONS

Question 1 :

Find the equation of a 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.

First let us find the point of intersection of the lines 4x + 7y − 3 = 0 and 2x − 3y  + 1 = 0

4x + 7y − 3 = 0 ------(1)

2x − 3y  + 1 = 0 ------(2)

3(1) + 7(2)

12x + 21y − 9 = 0

14x − 21y  + 7 = 0

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

26x - 2  =  0

x = 2/26

x = 1/13

Let us apply the value of x in (1), we get

4(1/13) + 7y - 3  =  0

7y - 3 + (4/13)  =  0

7y  =  3 - (4/13)

7y  =  (39-4)/13

7y  =  35/13

y  =  (35/13 x 7)

y  =  5/13

So, the point of intersection of the given lines is (1/13, 5/13). Since the required line is having equal intercepts, a = b

(x/a) + (y/b)  =  1

The line is passing through the point (1/13, 5/13)

((1/13)/a) + ((5/13)/a)  =  1

(1/13a) + (5/13a)  =  1

6/13a  =  1

13a  =  6

a  =  6/13

Equation of the line :

(x/(6/13)) + (y/(6/13))  =  1

(13x + 13y)/6  =  1

13x + 13y  =  6

13x + 13y - 6  =  0

Question 2 :

A person standing at a junction (crossing) of two straight paths represented by the equations 2x −3y + 4 = 0 and 3x + 4y −5 = 0 seek to reach the path whose equation is 6x −7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

The equation of the given lines are

2x −3y + 4 = 0  -----(1)

3x + 4y −5 = 0  -----(2)

6x −7y + 8 = 0   -----(3)

the person is standing at the junction of paths represented by the line (1) and (2)

By solving the (1) and (2), we get

4(1) + 3(2)

8x - 12y + 16  =  0

9x + 12y - 15  =  0

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

17x + 1  =  0

x  =  -1/17

By applying the value of x in (1), we get

2(-1/17) - 3y + 4  =  0

-3y  =  -4 + (2/17)

-3y  =  (-68 + 2)/17

-3y  =  - 66/17

y  =  66/17(3)  =  22/17

Thus the person is standing at the point (-1/17, 22/17)

The person can reach path (3) is the least time if he takes along the perpendicular the line to (3) from the point (-1/17, 22/17)

Slope of the line (3)

m  =  - coefficient of x/coefficient of y

m  =  -6/7

Equation of the line passing through the point (-1/17, 22/17) and having the slope -6/7

Slope of the required line  =  7/6

(y - y1)  =  m(x - x1)

(y - (22/17))  =  (7/6) (x + (1/17))

(17y - 22)/17  =  (7/6)(17x + 1)/17

6(17y - 22)  =  7(17x + 1)

102y - 132  =  119x + 7

119x + 102y + 132 - 7  =  0

119 x + 102y + 125  =  0

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