FIND EQUATION OF THE LINE FROM THE GRAPH

Find equation of the line from the following graph given :

Example 1  :

Solution :

Slope of a line  =  1/3

y – intercept(b)  =  2

Equation of a straight line y  =  mx + b

y  =  1/3x + 2

3y  =  x + 6

x – 3y + 6  =  0

So, the required equation is x – 3y + 6  =  0.

Example 2 :

Solution :

x – intercept form a  =  2

N – intercept form b  =  -1

By using x – intercept and y – intercept formula,

x/a + y/b  =  1

x/2 + N/(-1)  =  1

x/2 – N/1  =  1

- N  =  - 1/2x + 1

N  =  (1/2)x - 1

So, the required equation is N  =  (1/2)x - 1

Example 3 :

Solution  :

s – intercept form a  =  4

G – intercept form b  =  2

By using x – intercept and y – intercept formula,

x/a + y/b  =  1

S/4 + G/2  =  1

G/2  =  - S/4 + 1

G  =  - S/2 + 2

G  =  (-1/2)S + 2

So, the required equation is G  =  (-1/2)S + 2

Example 4 :

Solution :

In figure, the straight line passes through two points (4, -3) and (0, 2)

By using two points formula,

(y – y1)/(y2 – y1)  =  (x – x1)/(x2 – x1)

We have two points (4, -3) and (0, 2)

(4, -3)----->(x1, y1)

(0, 2)----->(x2, y2)

By applying the values, we get

(H+3)/(2+3)  =  (g–4)/(0–4)

(H+3)/5  =  (g–4)/(-4)

(-4) (H+3)  =  5 (g–4)

- 4H–12  =  5g–20

- 4H  =  5g–20+12

- 4H  =  5g–8

H  =  (-5/4)g + 2

So, the required equation is H  =  (-5/4)g + 2

Example 5 :

Solution :

In figure, the straight line passes through two points (10, 2) and (0, 1)

By using two points formula,

(y – y1)/(y2 – y1)  =  (x – x1)/(x2 – x1)

We have two points (10, 2) and (0, 1)

(10, 2)------>(x1, y1)

(0, 1)------>(x2, y2)

By applying the values, we get

(F–2)/(1–2)  =  (x–10)/(0-10)

(F–2)/(-1)  =  (x–10)/(-10)

(-10) (F–2)  =  (-1) (x–10)

-10F+20  =  - x+10

- 10F  =  - x+10–20

- 10F  =  -x–10

F  =  (1/10)x + 1

So, the required equation is F  =  (1/10)x + 1

Example 6 :

Solution :

In figure, the straight line passes through two points (6, -3) and (0, -2)

By using two points formula,

(y – y1)/(y2 – y1)  =  (x – x1)/(x2 – x1)

We have two points (6, -3) and (0, -2)

(6, -3)------>(x1, y1)

(0, -2)------>(x2, y2)

By applying the values, we get

(P+3)/(-2+3)  =  (t–6)/(0–6)

(P+3)/1  =  (t–6)/(-6)

(-6) (P+3)  =  t–6

- 6P–18  =  t–6

- 6P  =  t–6+18

- 6P  =  t+12

P  =  (-1/6)t - 2

So, the required equation is P  =  (-1/6)t – 2

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Nov 16, 24 08:15 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 71)

    Nov 16, 24 08:03 AM

    digitalsatmath56.png
    Digital SAT Math Problems and Solutions (Part - 71)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 70)

    Nov 15, 24 07:12 PM

    Digital SAT Math Problems and Solutions (Part - 70)

    Read More