We already learned how to represent a point and a line on a coordinate axis using cartesian coordinates. In this lesson, we will take up certain formulas concerning two points in the plane and also derive an equation of a line.
Consider point P lies on line segment AB.
P divides AB internally in ratio AP: PB. In the above figure, Point P divides AB internally in ratio 2 : 3.
Here we are going to find the coordinates of the point which divides internally the line joining two given points in a given ratio. Let A (x1, y1) and B (x2, y2) be the two given points and P (x, y) is a point on line AB which divides the line into ratio m1 : m2, then the coordinates of P are
\(x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}\\ y= \frac{m_1y_2 + m_2y_1}{m_1+m_2}\\\) |
What will be the coordinates of the mid-point?
Putting m1 = 1, m2 = 1, the coordinates of mid-point are:
\(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\)
Example: Find the coordinates of the point which divides the line joining the points (4 6) and (-4, 2) internally in the ratio 1:3.
Solution:
\(x = \frac{(1\times-4) + (3 \times4)}{1+3} = \frac{8}{4} = 2\)
\(y = \frac{(1\times2) + (3 \times6)}{1+3} = \frac{20}{4} = 5\)
Answer: (2, 5) are the coordinates of the point.
The slope (or gradient) of a straight line is the tangent of the angle which the part of the line above the x-axis makes with the positive direction of the x-axis.
The slope of a line is indicated by the letter 'm'.
AP is the portion of the line above the x-axis and \(\angle XAP = \theta\) then \(m = \tan\theta\). The angle is measured from the positive direction of the x-axis in an anticlockwise direction to the part of the line above the x-axis.
Slope of a line joining two points
Let A (x1, y1) and B (x2, y2) be two points. Let θ be the inclination of line AB with the x-axis then
SLOPE-INTERCEPT FORM OF A LINE
If a straight line l meets the x-axis at A and the y-axis at B then
(i) OA is called the intercept made by the line on the x-axis or simply x-intercept.
(ii) OB is called the intercept made by the line on the y-axis or simply y-intercept.
Let P be any point on a line l with coordinates (x, y). Line l makes an angle θ with X-axis and whose intercept OB on the y-axis is c.
Tan θ = m
The tangent of the angle is equal to the slope of the line.
The equation of a straight line with slope m and y-intercept c is y = mx + c.
ONE-POINT FORM OF A STRAIGHT LINE
The equation of a straight line passing through a given point (x1,y1) and given slope m is:
y - y1 = m ( x − x1)
TWO-POINTS FORM OF A STRAIGHT LINE
The equation of a straight line passing through two-point (x1,y1) and (x2,y2) and given slope m is:
\(\mathbf {y - y_1 = \frac {y_1 - y_2}{x_1 - x_2} ( x - x_1) }\)
Example: Find the equation of the straight line through the point (-1, -2) and having slope equal to 4/5.
Solution: The required equation is
\(y - (-2) = \frac{4}{5}{{x - (-1)}} \\ y + 2 = \frac{4}{5} (x + 1) \\ 5y + 10 = 4x + 4 \\ -4x + 5y = -6\)
Answer: -4x + 5y = -6 is the equation of the straight line.
Example: Determine the slope and y-intercept of y = 3x + 5.
Solution: comparing with y = mx + c, we get m = 3 and c = 5
Answer: Slope = 3, Intercept = 5
INTERCEPT FORM OF A STRAIGHT LINE
The equation of a straight line that cuts off intercept a and b from the axes is:
