Question

A straight line PQ cuts the x andy axis at M and N respectively. If the points A(-3,5) and B(4,7) lies on PQ , calculate
Gradient of PQ
Equation of PQ

Answer 1

Solution :-

Given that

• A straight line PQ cuts the x andy axis at M and N respectively.

• The points A(-3,5) and B(4,7) lies on PQ.

It implies,

• The gradient or slope of PQ is same as that of AB.

• The equation of PQ is same as that equation of AB.

We know,

Slope of line joining the points (a, b) and (c, d) is represented by m and is evaluated as

$\boxed{ \bf\:m= \dfrac{d - b}{c - a}}$

Here, points are A (- 3, 5) and B (4, 7).

Thus,

Gradient of line segment joining A and B is

$\rm :\longmapsto\: Gradient\: of \: AB \: = \: \dfrac{7 - 5}{4 - ( - 3)}$

$\rm :\longmapsto\:Gradient \: of \: AB \: = \: \dfrac{2}{4 + 3}$

$\rm :\longmapsto\:Gradient \: of \: AB \: = \: \dfrac{2}{7}$

Hence,

$\rm :\longmapsto\: Gradient\: of \: PQ \: = \: Gradient\: of \: AB$

$\bf\implies \:Gradient \: of \: PQ \: = \: \dfrac{2}{7}$

Now,

We know that

Equation of line passes through the point (a, b) and having gradient 'm' is given by

$\boxed{ \bf \: y - b \: = \: m(x - a)}$

Thus,

Equation of line AB passes through the point (- 3, 5) having slope 2/7 is

$\rm :\longmapsto\:y - 5 = \dfrac{2}{7}\bigg(x - ( - 3)\bigg)$

$\rm :\longmapsto\:y - 5 = \dfrac{2}{7}\bigg(x + 3\bigg)$

$\rm :\longmapsto\:7y - 35 = 2x + 6$

$\rm :\longmapsto\:2x - 7y + 41 = 0$

So,

Equation of line PQ is

$\bf :\longmapsto\:2x - 7y + 41 = 0$

Additional Information

Additional Information Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

• Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

• Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line

2. Point-slope form equation of line

• Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a)

3. Slope-intercept form equation of line

• Consider a line whose slope is m which cuts the Y-axis at a distance ‘a’ from the origin. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a.

4. Intercept Form of Line

• Consider a line L having x– intercept a and y– intercept b, then the line passes through  X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1.

5. Normal form of Line

• Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, the equation of line is given by x cosβ + y sinβ = p.

Answer 2

Answer:

Step-by-step explanation: