In this diagram, O is the origin and PQRS is a parallelogram. The equation of the straight line QR is $2y = -x + 13$ P (0,3) R (9,2) S a) Find the equation of the straight line RS b) Find the y-intercept of the straight line RS
Added by Jeremy B.
Close
Step 1
To find the y-intercept of a line, we set x = 0 in the equation of the line and solve for y. When x = 0, the equation 2y = -x + 13 becomes 2y = 0 + 13, which simplifies to 2y = 13. Dividing both sides of the equation by 2, we get y = 13/2. So, the y-intercept of Show more…
Show all steps
Your feedback will help us improve your experience
Jason Jacob and 50 other Geometry educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Express u as a linearcombination $\mathbf{u}=r \mathbf{v}+s \mathbf{w}$ Then sketch $\mathbf{u}, \mathbf{v}, \mathbf{w},$ and the parallelogram formed by $r \mathbf{v}$ and sw. $\mathbf{u}=\langle 3,-1\rangle ; \quad \mathbf{v}=\langle 2,1\rangle, \mathbf{w}=\langle 1,3\rangle$
VECTOR GEOMETRY
Vectors in the Plane
All lines are in the $(x, y)$ plane. Write the equation of the straight line through $(2,-3)$ with slope $\frac{3}{4}$, in the paramctric form $\mathbf{r}=r_{0}+\mathbf{A}$.
LINEAR EQUATIONS; VECTORS, MATRICES, AND DETERMINANTS
Lines and planes
Find the points of intersection of the given line and plane. $$r=\left(\begin{array}{l}2 \\4 \\0\end{array}\right)+\mu\left(\begin{array}{c}0 \\-3 \\3\end{array}\right) \text { and } 4 x-2 y+3 z-30=0$$
Vectors, Lines and Planes
Planes
Recommended Textbooks
Geometry A Common Core Curriculum
Geometry
Watch the video solution with this free unlock.
EMAIL
PASSWORD