color(brown)("The slope (gradient) is stated as a single value so this is a straight line graph")The slope (gradient) is stated as a single value so this is a straight line graph
color(blue)("Determine the equation of the line")Determine the equation of the line
Gradient is m=("change in y")/("change in x reading left to right")m=change in ychange in x reading left to right
m=("change in y")/("change in x reading left to right") = (-1)/2m=change in ychange in x reading left to right=−12
So the y value becomes less as you read left to right. The slope is down.
Let the given point be P_1->(x_1,y_1)=(-3,-5)P1→(x1,y1)=(−3,−5)
Using the standardised equation format y=mx+cy=mx+c
We have by substituting the values for P_1P1
y_1=mx_1+c" "->" "-5=(-1/2)(-3)+cy1=mx1+c → −5=(−12)(−3)+c
" "-5=+3/2+c −5=+32+c
Subtract 3/232 from both sides to get cc on its own.
" "-5-3/2=0+c −5−32=0+c
" "-13/2=c −132=c
So the equation of the line that passes through the point (-3,-5) is:
y=-1/2x - 13/2y=−12x−132
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color(blue)("Determine the value of the y-intercept")Determine the value of the y-intercept
The y-intercept is the value of the constant c=-13/2c=−132
ie: set x=0x=0
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color(blue)("Determine the value of the x-intercept")Determine the value of the x-intercept
Set y=0y=0
y=-1/2x-13/2" "->" "0=-1/2x-13/2y=−12x−132 → 0=−12x−132
1/2x=-13/212x=−132
x=-26/2 = -13x=−262=−13
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Mark your points on the x and y axis and draw a straight line through them. Also show the given point.
![Tony B]()
Tony B
