**Equations of Lines Advanced Questions:**

Here we are going to see an example problem on the topic coordinate geometry.

**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.

**Solution :**

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.

**Solution :**

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 - y_{1}) = m(x - x_{1})

(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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