The straight line that passes through the points $(4, v)$ and $(8,1)$ has gradient $\frac{5}{7}$. What is the value of $v$ ? Give your answer as an integer or as a fraction in its simplest form.
Answer (2)
Use the gradient formula: 7 5 = 8 − 4 1 − v .
Simplify the equation: 7 5 = 4 1 − v .
Solve for v : v = 1 − 7 20 .
Find the value of v : − 7 13 .
Explanation
Understanding the Problem We are given two points ( 4 , v ) and ( 8 , 1 ) on a straight line, and we know that the gradient of this line is 7 5 . Our goal is to find the value of v .
Using the Gradient Formula The gradient of a line passing through two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by the formula: gradient = x 2 − x 1 y 2 − y 1 In our case, ( x 1 , y 1 ) = ( 4 , v ) and ( x 2 , y 2 ) = ( 8 , 1 ) , and the gradient is 7 5 .
Substituting the Values Substituting the given values into the gradient formula, we get: 7 5 = 8 − 4 1 − v 7 5 = 4 1 − v
Solving for v Now, we solve for v . Multiply both sides of the equation by 4: 4 × 7 5 = 1 − v 7 20 = 1 − v
Finding the Value of v Add v to both sides and subtract 7 20 from both sides: v = 1 − 7 20 To subtract the fraction, we need a common denominator, which is 7. So, we rewrite 1 as 7 7 :
v = 7 7 − 7 20 v = 7 7 − 20 v = 7 − 13
Final Answer Therefore, the value of v is − 7 13 .
Examples
Imagine you are designing a ramp for a skateboard park. You know two points the ramp must pass through and the desired steepness (gradient). Using the gradient formula, you can determine the height of the ramp at one of those points, ensuring the ramp is safe and fun for skateboarders. This problem demonstrates how coordinate geometry and gradients are used in real-world design and engineering.
The value of v is − 7 13 , found by using the gradient formula between the points ( 4 , v ) and ( 8 , 1 ) with a given gradient of 7 5 . After plugging in the values and simplifying, we determined v through algebraic manipulation.
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