just need some help with these questions! thanks!!! along with a small explanation would be great!
1.Find the equation of the line through (7,1) which meets the x axis at x=3. Enter your solution in the form y=mx+b.
2.Find the equation of the line which meets the x-axis at x=4 and the y-axis at y=4. Enter your solution in the form y=mx+b.
3.Find where the line which passes through the two points (1,2) and (3,5) intersects the line through (2,1) and (9,-9).
4.Find the equation of the line x=1t+6 and y=3-5t in the form y=mx+b.
thanks!
1.Find the equation of the line through (7,1) which meets the x axis at x=3. Enter your solution in the form y=mx+b.
2.Find the equation of the line which meets the x-axis at x=4 and the y-axis at y=4. Enter your solution in the form y=mx+b.
3.Find where the line which passes through the two points (1,2) and (3,5) intersects the line through (2,1) and (9,-9).
4.Find the equation of the line x=1t+6 and y=3-5t in the form y=mx+b.
thanks!
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The simple answer for 1,2 is that you are given two points...and so you can construct a line from them....one of the axioms of geometry actually...anyhow
so for question one, one cord is (7,1)...the second is (3,0) <--- convince yourself that this is true graphically if needed...and so you can find m by the following....m=(y1 - y2)/(x1-x2)...it doesnt matter which x and y you choose for either. once you have m...then write out your equation so that one of your points fills the value for the x and y. Also plug your m in. From here you will want to simply solve for b. take your m and your b, then plug them BACK into the equation with x any y....so example y=#x + #
For number 3 you will want to do the steps listed above to get the equations of the two corresponding lines...from here you will have two equations y1 = m1x1 + b1, and y2 = m2x2 + b2. Now the thought is that the if the two lines do indeed meet, then they will have an x and a y value that are the exact same....so they would be equal in that point, and so any time you have equality like this you can set them equal to each other....so m1x + b1 = m2x + b2....then solve for x...now this x value is the value where they meet...all you need now is to plug in the x into one of the equations and solve for the corresponding y value....and done!...you answer should be : (x,y) format
for the last one...your equation would have all numbers and an x and a y...if you solve BOTH equations for t...IE t = (stuff of x ) and t = (stuff of y), then again set them equal to each other....from here simply solve for y and then done.
so for question one, one cord is (7,1)...the second is (3,0) <--- convince yourself that this is true graphically if needed...and so you can find m by the following....m=(y1 - y2)/(x1-x2)...it doesnt matter which x and y you choose for either. once you have m...then write out your equation so that one of your points fills the value for the x and y. Also plug your m in. From here you will want to simply solve for b. take your m and your b, then plug them BACK into the equation with x any y....so example y=#x + #
For number 3 you will want to do the steps listed above to get the equations of the two corresponding lines...from here you will have two equations y1 = m1x1 + b1, and y2 = m2x2 + b2. Now the thought is that the if the two lines do indeed meet, then they will have an x and a y value that are the exact same....so they would be equal in that point, and so any time you have equality like this you can set them equal to each other....so m1x + b1 = m2x + b2....then solve for x...now this x value is the value where they meet...all you need now is to plug in the x into one of the equations and solve for the corresponding y value....and done!...you answer should be : (x,y) format
for the last one...your equation would have all numbers and an x and a y...if you solve BOTH equations for t...IE t = (stuff of x ) and t = (stuff of y), then again set them equal to each other....from here simply solve for y and then done.