Back to: MATHEMATICS SS3

**Welcome to class! **

In today’s class, we will be talking about coordinate geometry of a straight line. Enjoy the class!

**Coordinate Geometry of a Straight Line**

Any straight line has an equation of the form

y = mx + c

where m, the gradient, is the height through which the line rises in one-unit step in the horizontal direction

c, the intercept, is the y-coordinate of the point of intersection between the line and the y-axis. This is shown in the figure below.

The straight line, y = mx + c

If we know the gradient m of a straight line with unknown intercept c, and the coordinates (x_{1} , y_{1}) of a point through which it passes, then we know that

y_{1 }= mx_{1 }+ c

and therefore

c = y_{1 }− mx_{1}

If we substitute into

y = mx + c

we obtain

y = mx− mx_{1 }+ y_{1}

which we can rearrange to give

y− y_{1 }= m(x − x_{1})

If we know two points (x_{1} , y_{1}) and (x_{2} , y_{2}) through which passes a line with unknown gradient m and intercept c, then

y_{1} = mx_{1 }+ c ,

y_{2} = mx_{2 }+ c

Subtracting the first equation from the second gives

y_{2 }− y_{1 }= m(x_{2 }− x_{1})

and therefore

$m=(x2\u2013x1)/(y2\u2013y1)$

The equation of the line is therefore

$y-y1=(y2\u2013y1)/(x2\u2013x1)(x-x1)$

**The midpoint of the line joining two points**

Once we know the coordinates of two points on a straight line, we can find the mid-point of the line.

**Distance between two points**

The distance formula is derived from the Pythagoras theorem. To find the distance between two points, if A is (x_{1} , y_{1}) and B is (x_{2} , y_{2}), all that you need to do is use the coordinates of these ordered pairs.

The distance between two points is given by:

$Distance=\surd (\u3016(x2\u2013x1){\u3017}^{2}+\u3016(y2\u2013y1\left){\u3017}^{2}\right)$

In our next class, we will be talking more about the **Coordinate Geometry of straight line**. We hope you enjoyed the class.

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prosperakpe510the formula for finding the midpoint of a straight line is what?