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Find the equations (in terms of x) of the line through the points (-2,-3) and (3,-5)

User Ashutosh A
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1 Answer

19 votes
19 votes

The general equation of a line passing through two points (xb₁,y₁)Pxb₂,y₂) is expressed as


\begin{gathered} y-y_1=m(x-x_1) \\ \text{where} \\ m\Rightarrow slope\text{ of the line, expr}essed\text{ as }m\text{ = }(y_2-y_1)/(x_2-x_1) \\ (x_1,y_1)\Rightarrow coordinate_{}\text{ of point P} \\ (x_2,y_2)\Rightarrow coordinate_{}\text{ of point Q} \end{gathered}

Given that the coordinates of the two points are (-2, -3) and (3, -5), we have


\begin{gathered} (x_1,y_1)\Rightarrow(-2,\text{ -3)} \\ (x_2,y_2)\Rightarrow(3,\text{ -5)} \end{gathered}

Step 1:

Evaluate the slope o the line.

The slope is thus evaluated as


\begin{gathered} m\text{ = = }(y_2-y_1)/(x_2-x_1) \\ \text{ = }\frac{\text{-5-(-3)}}{3-(-2)} \\ =(-5+3)/(3+2) \\ \Rightarrow m\text{ = -}(2)/(5) \end{gathered}

Step 2:

Substitute the values of x₁,

Thus, we have


\begin{gathered} y-y_1=m(x-x_1) \\ x_1=-2 \\ y_1=-3 \\ m\text{ =- }(2)/(5) \\ \text{thus,} \\ y-(-3)\text{ = -}(2)/(5)(x-(-2)) \\ y+3\text{ =- }(2)/(5)(x+2) \end{gathered}

Step 3:

Make .


\begin{gathered} y+3\text{ =- }(2)/(5)(x+2) \\ \text{Multiply both sides of the equation by 5 } \\ 5(y+3)\text{ = -2(x+2)} \\ \text{open brackets} \\ 5y\text{ + 15 =- 2x - 4} \\ \Rightarrow5y\text{ =- 2x - 4 -15} \\ 5y\text{ = -2x-1}9 \\ \text{divide both sides of the equation by the coefficient of y, which is 5.} \\ \text{thus,} \\ (5y)/(5)=\frac{-\text{2x-1}9}{5} \\ \Rightarrow y\text{ =- }(2)/(5)x\text{ - }(19)/(5) \end{gathered}

Hence, the equation of the line is


y\text{ = -}(2)/(5)x\text{ - }(19)/(5)

y₁ and m into the general equation of the line.

User Reticent
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