EQUATION OF THE LINE PASSING THROUGH THE POINT AND SUM OF INTERCEPT

Example 1 :

Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3 : 2.

Solution :

Since the intercepts are in the ratio 3 : 2,

x-intercept(a) = 3k and y-intercept(b) = 2k

(x/a) + (y/b)  =  1

(x/3k) + (y/2k)  =  1

The straight line is passing through the point (3, 4)

(3/3k) + (4/2k) = 1

(1/k) + (2/k) = 1

(1 + 2)/k = 1 ==>  3/k = 1 ==> k = 3

a = 3k = 9, b = 2k = 6

Equation of the line :

(x/a) + (y/b) = 1

(x/9) + (y/6) = 1

(2x + 3y)/18 = 1

2x + 3y = 18

2x + 3y - 18 = 0

Example 2 :

Find the equation of the straight lines passing through the point (2, 2) and the sum of the intercepts is 9.

Solution :

Sum of intercepts = 9

a + b = 9

a = 9 - b 

(x/a) + (y/b)  =  1

(x/(9-b)) + (y/b)  =  1

The straight line is passing through the point (2, 2)

(2/(9-b)) + (2/b)  =  1

[2b + 2(9 - b)]/[b(9-b)]  =  1

(2b + 18 - 2b)/(9b-b²)  =  1

18 = 9 b - b²

b² - 9 b - 18 = 0

(b - 3) (b - 6) = 0

Example 3 :

Find a point on the x-axis, which is equidistance from the points (7, 6) and (3, 4).

Solution :

Let a be the point on the x-axis.

(a, 0)

Distance between the points (a, 0) and (7, 6) = distance between the points (a, 0) and (3, 4)

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

√(7 - a)² + (6 - 0)² = √(3 - a)² + (4 - 0)²

(7 - a)² + 36 = (3 - a)² + 16

49 - 14a + a² + 36 = 9 - 6a + a² + 16

85 - 14a = 25 - 6a

-14a + 6a = 25 - 85

-8a = 60

a = 60/(-8)

a = -15/2

So, the require x-intercept is (-15/2, 0).

Example 4 :

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points (0,−4) and (8, 0).

Solution :

Midpoint of the line joining the points (0, -4) and (8, 0)

Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

= (0 + 8)/2, (-4 + 8)/2 

= 8/2, 4/2

= (4, 2)

Slope of the line passes trough the points (0, 0) and (4, 2).

Slope = (2 - 0) / (4 - 0)

= 2/4

= 1/2

So, the required slope is 1/2.

Example 5 :

Find the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3)

Solution :

x-intercept is (2, 0) and y-intercept is (0, 3)

The equation will be in the form of y = mx + b

The line passes through the point (2, 0), we get

0 = m(2) + b

0 = 2m + b

2m + b = 0  ---------(1)

The line passes through the point (0, 3), we get

3 = m(0) + b

3 = 0 + b

b = 3

Applying the value of b in (1), we get

2m + 3 = 0

2m = -3

m = -3/2

Applying the values of slope and y-intercept, we get

y = (-3/2)x + 3

Find the equation of the straight line which passes through the point (3, 4) and the sum of its intercepts on the axes is 14.

Solution :

Sum of intercepts = 14

a + b = 14

a = 14 - b 

(x/a) + (y/b)  =  1

(x/(14-b)) + (y/b)  =  1

The straight line is passing through the point (3, 4)

(3/(14 - b)) + (4/b)  =  1

[3b + 4(14 - b)]/[b(14 - b)]  =  1

(3b + 56 - 4b)/(14b - b²)  =  1

-b + 56 = 14b - b²

b² - b - 14b + 56 = 0

b² - 15b + 56 = 0

(b - 8)(b - 7) = 0

b = 7 and b = 8

a = 14 - b 

When b = 7, a = 14 - 7 ==> 7

When b = 8, a = 14 - 8 ==> 6

Equation of line when a = 7 and b = 7

x/7 + y/7 = 1

x + y = 7

Equation of line when a = 6 and b = 8

x/6 + y/8 = 1

4x + 3y = 24

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