Dear Secondary Math students, Math Lobby will be going through another section in coordinate geometry which is how to find an equation of a straight line. We will be teaching you how to use the famous y = mx + c equation. Let’s begin!

__In this note, you will learn:__

__In this note, you will learn:__

### · __How to find the equation of a straight line in a graph in the form of y = mx + c__

__How to find the equation of a straight line in a graph in the form of y = mx + c__

A typical type of question involves finding the equation of a line in a graph, formed in the manner of: **y = mx + c, where m = gradient of the line, and c = y-intercept of the line**.

Now, how do we solve these types of questions instead? Let’s find out!

Examination questions will usually prompt students to find the gradient of the graph first, which by mathematical definition in layman terms, is a number which describes both the direction and the steepness of the line.

It is defined with the equation: **Gradient = (y1 – y2) / (x1 – x2)**, where (x1,y1) and (x2,y2) are coordinates that lies on the line.

Thereafter, you can substitute the value of the gradient into **m **and any x and y-coordinates that lies on the line to find the value of the y-intercept of the line, **c**. For more information on ** how to find the gradient of a straight line**, do check out our article above!

Let’s apply this knowledge on a question:

**Given that we have a line which has the points C (5,6) and D (7,9), find the equation of the line.**

First of all, let’s take a look at what we are provided with. We are given two sets of coordinates, **C** and **D**, which lies on the line of the equation which we are interested to find.

Hence, we are able to calculate the gradient of the line! Let’s substitute the coordinates provided into the formula:

Gradient = (y1 – y2) / (x1 – x2)

*Note: The order of which coordinates you use to be x1, y1 or x2, y2 doesn’t matter. Just be sure that you do not make any careless mistakes and substitute the wrong values into the formula!

m = (9 – 6) / (7 – 5)

= 3 / 2

Next, we are able to substitute m = 3 / 2 and any set of coordinates that lies on the line. In this case, we will use point C, which has the coordinates of (5,6).

y = mx + c

Substituting coordinates (5,6) and m = 3 / 2 into the equation,

(6) = (3/2) (5) + c

Therefore, c = 6 – (15 / 2)

= -3 / 2

Hence, the equation of the line is **y = (3/2) x – 3/2**

Now to double-check your answers, just simply substitute any x or y-value that you know lies on the line and see if you get the coordinate point of the other axis! In this case, we will use the x-coordinate of point D,

Substituting x = 7 into the equation **y = (3/2) x – 3/2**,

y = (3/2) (7) – 3/2

Therefore, y = 9

This shows that the line, **y = (3/2) x – 3/2**, is valid for both points C and D.

And that is how you find the equation of a line in a graph!

Now, let’s get your brains working!

Question:

Given that we have a quadratic curve with the equation y = x2 + 8x - 9,

i) Sketch the quadratic curve on a graph paper, showing the x, y-intercept and the turning points of the equation clearly.

ii) Draw a line of tangent to the curve where x = -3, and find the gradient of the line of tangent.

iii) Find the equation of the line of tangent.

Answers:

i)

ii)

For part ii), we were told to find the gradient of the line of tangent. To do so, we just have to use any two points that lies on the line, which in this case we are using the coordinates (20, 14) and (2, -16). *Note: In order to yield a gradient value of a higher accuracy, try to find points that are spaced out as far as possible.

Calculation for part ii):

Gradient = (y1 – y2) / (x1 – x2)

= [14 – (-16)] / (20 – 2)

= 30 / 18

= 1.67 (corrected to 3 s.f)

iii)

Since we have a set of coordinates that lies on the line, and we have the gradient of the line which we found back in part ii), we can simply substitute the values into the equation y = mx + c to get the value of the y-intercept and hence the equation!

y = mx + c

Substituting coordinates (20, 14) and m = 30 / 18 into the equation y = mx + c,

(14) = (30/18) (20) + c

Therefore, c = -58 / 3

Hence, the equation of the line of tangent is y = (30/18) x – 58/3

And that’s all we have for today, students! ** Math Lobby** hopes that all of you have had a clear understanding on how to solve for the equation of a line in a graph, and for any questions, feel free to contact us on Facebook, Instagram or Website! And as always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!

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