( $3$ ) A straight line has the equation $a x+b y=5$ , where $a$ and $b$ are constants. $\newline$ It passes through the points $(3,2)$ and $(1,-1)$ . Find $\newline$ (a) the value of $a$ and of $b$ , $\newline$ (b) the gradient of the line.

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Find a missing coordinate using slope

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Q. (

3

) A straight line has the equation

a x+b y=5

, where

a

and

b

are constants.

\newline

It passes through the points

(3,2)

and

(1,-1)

. Find

\newline

(a) the value of

a

and of

b

\newline

(b) the gradient of the line.

Substitute Coordinates: To find the values of $a$ and $b$ , we can substitute the coordinates of the given points into the equation $ax+by=5$ . Let's start with the point $(3,2)$ . Substitute $x=3$ and $y=2$ into the equation: $a(3) + b(2) = 5$ $3a + 2b = 5$
Equations with Two Variables: Now, let's substitute the coordinates of the second point $(1,-1)$ into the equation: $\newline$ $a(1) + b(-1) = 5$ $\newline$ $a - b = 5$
Elimination Method: We now have a system of two equations with two variables: $\newline$ $3a + 2b = 5$ $\newline$ $a - b = 5$ $\newline$ We can solve this system using the method of substitution or elimination. Let's use elimination. $\newline$ First, we can multiply the second equation by $2$ to align the coefficients of $b$ : $\newline$ $2(a - b) = 2(5)$ $\newline$ $2a - 2b = 10$
Solve for $a$ : Now we have the system: $\newline$ $3a + 2b = 5$ $\newline$ $2a - 2b = 10$ $\newline$ We can add these two equations together to eliminate $b$ : $\newline$ $(3a + 2b) + (2a - 2b) = 5 + 10$ $\newline$ $3a + 2a = 15$ $\newline$ $5a = 15$
Substitute for $b$ : Divide both sides by $5$ to find the value of $a$ : $\frac{5a}{5} = \frac{15}{5}$ $a = 3$
Find Gradient: Now that we have the value of $a$ , we can substitute it back into one of the original equations to find $b$ . Let's use the second equation: $a - b = 5$ $3 - b = 5$
Find Gradient: Now that we have the value of $a$ , we can substitute it back into one of the original equations to find $b$ . Let's use the second equation: $\newline$ $a - b = 5$ $\newline$ $3 - b = 5$ Subtract $3$ from both sides to solve for $b$ : $\newline$ $3 - b - 3 = 5 - 3$ $\newline$ $-b = 2$ $\newline$ $b = -2$
Find Gradient: Now that we have the value of $a$ , we can substitute it back into one of the original equations to find $b$ . Let's use the second equation: $\newline$ $a - b = 5$ $\newline$ $3 - b = 5$ Subtract $3$ from both sides to solve for $b$ : $\newline$ $3 - b - 3 = 5 - 3$ $\newline$ $-b = 2$ $\newline$ $b = -2$ We have found the values of $a$ and $b$ : $\newline$ $b$ $1$ $\newline$ $b = -2$ $\newline$ Now we need to find the gradient of the line. $\newline$ The gradient (slope) of a line is given by the change in $b$ $3$ divided by the change in $b$ $4$ between two points on the line. $\newline$ Using the points $b$ $5$ and $b$ $6$ , we can calculate the gradient as follows: $\newline$ Gradient = $b$ $7$ $\newline$ Gradient = $b$ $8$
Find Gradient: Now that we have the value of $a$ , we can substitute it back into one of the original equations to find $b$ . Let's use the second equation: $\newline$ $a - b = 5$ $\newline$ $3 - b = 5$ Subtract $3$ from both sides to solve for $b$ : $\newline$ $3 - b - 3 = 5 - 3$ $\newline$ $-b = 2$ $\newline$ $b = -2$ We have found the values of $a$ and $b$ : $\newline$ $b$ $1$ $\newline$ $b = -2$ $\newline$ Now we need to find the gradient of the line. $\newline$ The gradient (slope) of a line is given by the change in $b$ $3$ divided by the change in $b$ $4$ between two points on the line. $\newline$ Using the points $b$ $5$ and $b$ $6$ , we can calculate the gradient as follows: $\newline$ Gradient = $b$ $7$ $\newline$ Gradient = $b$ $8$ Calculate the gradient: $\newline$ Gradient = $b$ $8$ $\newline$ Gradient = $a - b = 5$ $0$ $\newline$ Gradient = $a - b = 5$ $1$