Question

Find the equation of a straight line passing through a point 3,-2 and cuts of positive intercepts on the x-axis and y-axis which are in ratio4:3

Answer 1

Question:

Find the equation of a straight line passing through a point (3,-2) and cuts off positive intercepts on the x-axis and y-axis which are in ratio 4:3.

Answer:

The equation of the required straight line :

• In general form is ; 3x + 4y - 1 = 0 .

• In x,y-intercept form is ; x/(1/3) + y/(1/4) = 1 .

Note:

• If any straight line (or any other curve) passes through a given point, then the coordinates of the given point will satisfy the equation of that straight line (or that curve).

• The equation of a straight line in general form is given by ; Ax + By + C = 0.

• The equation of a straight line in slope, y-intercept form is given by ; y = mx + c , where , m is the slope and c is the y-intercept of the straight line.

• The equation of a straight line in x, y-intercept form is given by; x/a + b/y = 1 , where a is the x-intercept and b is the y-intercept.

Also, the point of interception on x-axis is (a,0) and the point of interception on y-axis is (0,b).

Solution:

It is given that;

The required straight line cuts off positive intercepts on the x-axis and y-axis which are in ratio 4:3. ie; a:b = 4:3.

Thus;

Let a = 4p and b = 3p.

Therefore,

The equation of the required straight line will be given by;

=> x/a + y/b = 1

=> x/4p + y/3p = 1

=> (3x + 4y)/12p = 1

=> 3x + 4y = 12p ----------(1)

Also,

It is given that, the required straight line passes through the given point (3,-2).

Thus;

The coordinates of the given point (ie; x = 3 and

y = -2) will satisfy the equation of the required straight line (ie; eq-1).

Now,

Putting x = 3 and y = -2 in eq-1 , we get;

=> 3x + 4y = 12p

=> 3•3 + 4•(-2) = 12p

=> 9 - 8 = 12p

=> 1 = 12p

=> p = 1/12

Now,

Putting p = 1/12 in eq-1 , we get;

=> 3x + 4y = 12p

=> 3x + 4y = 12•(1/12)

=> 3x + 4y = 12/12

=> 3x + 4y = 1

=> 3x + 4y - 1 = 0

This is the equation of the required line in general form.

Also,

=> 3x + 4y = 1

=> x/(1/3) + y/(1/4) = 1

This is the equation of the required straight line in x, y-intercept form.

Where,

x-intercept = 1/3

y-intercept = 1/4

Also;

Point of interception on x-axis is (1/3,0).

Point of interception on y-axis is (0,1/4).

Hence,

The equation of the required straight line in general form is ;

3x + 4y - 1 = 0 .

The equation of the required straight line in

x, y-intercept form is ;

x/(1/3) + y/(1/4) = 1 .