1. Consider the line passing through the points (1.3) and (4,9). a) What is the slope of the line ? b) What is the equation of the line ? c) Find the coordinates of the point at which the line cuts the x-axis ?
a.) Slope of the line = [tex]\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Slope of the line AB = [tex]\dfrac{9-3}{4-1} = \dfrac{6}{3}= \dfrac{2}{1} \:\:\text{or}\:\:2[/tex]
b.) Equation of the line – [tex]y – y_1=m(x – x_1)[/tex]
y = y coordinate of second point
m = slope
x = x coordinate of second point
=> y – 3=2(x-1)
=> y – 3 = 2x -2
=> y = 2x – 2 + 3
=> y = 2x + 1
c.) It is given in the question that we have to find the coordinates of the point at which the line cuts the x-axis. This indicates that y- coordinate of that point is 0 because it cuts x – axis.
Answer :-
a.) Slope of the line = [tex]\dfrac{2}{1} \:\:\text{or}\:\:2[/tex]
b.) Equation of the line – y = 2x + 1
c.) The coordinates of the point at which the line cuts the x-axis is (-0.5, 0)
Solution :-
Let the points be A(1,3) and B(4,9).
Here,
[tex]x_1= 1,\:\:\:\:x_2=4\\y_1 = 3,\:\:\:\:y_2=9[/tex]
a.) Slope of the line = [tex]\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Slope of the line AB = [tex]\dfrac{9-3}{4-1} = \dfrac{6}{3}= \dfrac{2}{1} \:\:\text{or}\:\:2[/tex]
b.) Equation of the line – [tex]y – y_1=m(x – x_1)[/tex]
y = y coordinate of second point
m = slope
x = x coordinate of second point
=> y – 3=2(x-1)
=> y – 3 = 2x -2
=> y = 2x – 2 + 3
=> y = 2x + 1
c.) It is given in the question that we have to find the coordinates of the point at which the line cuts the x-axis. This indicates that y- coordinate of that point is 0 because it cuts x – axis.
=> y = 2x + 1
=> 0 = 2x + 1
=> -1 = 2x
=> [tex]\dfrac{-1}{2} =x[/tex]
=> -0.5 = x
The point is – (-0.5, 0)