Solveeit Logo

Question

Question: How do you find the slope of \(\left( 0,0 \right), \left( 3,5 \right)\)?...

How do you find the slope of (0,0),(3,5)\left( 0,0 \right), \left( 3,5 \right)?

Explanation

Solution

The slope of a line passing through the points (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) is given by the formula m=y2y1x2x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}. Then by substituting the values of points in the formula and simplifying the obtained equation we will get the desired answer.

Complete step-by-step solution:
We have been given the points (0,0),(3,5)\left( 0,0 \right), \left( 3,5 \right).
We have to find the slope of a line passing through the given points.
We know that slope of a line is the ratio of the difference of y-coordinate to x-coordinate for the given two points. The value of slope may be positive, negative or zero.
Now, we know that the slope of a line passing through the points (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) is given by the formula m=y2y1x2x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}. When one point is (0,0)\left( 0,0 \right), it means the line starts from the origin.
Now, we have x1=0,y1=0,x2=3,y2=5{{x}_{1}}=0,{{y}_{1}}=0,{{x}_{2}}=3,{{y}_{2}}=5
Now, substituting the values in the formula we will get
5030\Rightarrow \dfrac{5-0}{3-0}
Now, simplifying the above obtained equation we will get
53\Rightarrow \dfrac{5}{3}
Hence we get the slope of a line passing through the points (0,0),(3,5)\left( 0,0 \right), \left( 3,5 \right) is 53\dfrac{5}{3}.

Note: The equation of slope-intercept form of the line is given as y=mx+cy=mx+c, where m is the slope of the line and c is the y-intercept of the line. We can also find the equation of a line by using the formula yy1xx1=y1y2x1x2\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} or (yy2)=(y2y1x2x1)(xx2)\left( y-{{y}_{2}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{2}} \right) passing through the points (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right).