need help with a set of problems

[SurnameLong]

A line has slope 3/4 and goes through the point (2,1). (a) Find two points on the line, each 5 units from (2,1). (b) Find two points on the line, each 8 units from (2,1).

Find the third vertex of an isosceles triangle whose base is the line joining (-2,-1) and (6,3), if the altitude on this base is square root 5. (two answers)

A Line L₁ has slope 2. Lines L₂ and L₃ have angles of inclination 45 degrees more, and 45 degrees less, respectively, than that of L₁. Find the slopes of L₂ and L₃.

 

[stapel]

A line has slope 3/4 and goes through the point (2,1). (a) Find two points on the line, each 5 units from (2,1). (b) Find two points on the line, each 8 units from (2,1).

What did you get as the equation of the line? How did you then use this information along with the Distance Formula?

Find the third vertex of an isosceles triangle whose base is the line joining (-2,-1) and (6,3), if the altitude on this base is square root 5. (two answers)

You plotted the points. You plugged them into the Distance Formula to get the lengths of the sides. And... then what?

A Line L₁ has slope 2. Lines L₂ and L₃ have angles of inclination 45 degrees more, and 45 degrees less, respectively, than that of L₁. Find the slopes of L₂ and L₃.

What formulas or techniques did they give you for doing this sort of exercise?

Please be complete. Thank you!

 

[SurnameLong]

Partially worked. Need help.

(1) A line has slope 3/4 and goes through the point (2,1). (a) Find two points on the line, each 5 units from (2,1). (b) Find two points on the line, each 8 units from (2,1)

I'm having trouble with (b). The equation of the line is y=3/4x -1/2
√((2-x)² + (1-y)²)=8, I multiplied and squared to get x²-4x +y²-2y=59. I then tried subbing in 3/4x -1/2 for the value of y. The answers are (42/5,29/5), (-22/5,-19/5), yet I have no idea as to how these answers were obtained. I obtained (a) by setting (1-y)/(2-x)=3/4 and (y-1)/(x-2)=3/4 to obtain (-2,-2) and (6,4). These points just happened to be 5 units away when subbed into the distance formula.

(2) Find the third vertex of an isosceles triangle whose base is the line joining (-2,-1) and (6,3), if the altitude on this base is square root 5. (two answers)

I found the midpoint between (-2,-1) and (6,3) to be (2,1). a²+b²=c², √20²+√5²=c². I found c to be 5 and put that in as D in the distance formula. From here, I know not how to proceed. The answer is (3,-1) or (1,3).

(3) A Line L₁ has slope 2. Lines L₂ and L₃ have angles of inclination 45 degrees more, and 45 degrees less, respectively, than that of L₁. Find the slopes of L₂ and L₃.

I don't know how to find the angle of inclination for the slope of 2.

 

[ksdhart]

For all of these problems, I strongly recommend drawing them out on graph paper. It will make the process so much easier.

A line has slope 3/4 and goes through the point (2,1). (a) Find two points on the line, each 5 units from (2,1). (b) Find two points on the line, each 8 units from (2,1).

For #1, you can solve part a and b in exactly the same way. The work you outlined for part b is exactly what I would do to solve this problem. You just seem to have stopped about an inch short of crossing the goal. There's one more quick step and then you'll know why the answers are what they are.

You have an equation x²-4x +y²-2y=59. When you plug in y=3x/4 - 1/2, what do you get? It is a quadratic equation, of the form ax^2 + bx + c = 0. Now solve that and you'll get the two values of x of 42/5 and -22/5. Then plug those x values into your equation to get the corresponding y values.

Then you can do the exact same process for part a, except you'll start with the distance formula equaling 5.

(2) Find the third vertex of an isosceles triangle whose base is the line joining (-2,-1) and (6,3), if the altitude on this base is square root 5. (two answers)

So you know two of the vertices of your triangle that form the base. You are correct in your first step of finding the mid-point of the base, but your next step doesn't look right to me. Use the distance formula to plot the circle defined by any point exactly sqrt(5) away from (2,1). That circle represents all possible locations of the third vertex.

You also know that it's an isosceles triangle, so two of the sides must be equal. Well, can you find some points which are equidistant from the two vertices you do know? Try and solve the problem from there.

(3) A Line L₁ has slope 2. Lines L₂ and L₃ have angles of inclination 45 degrees more, and 45 degrees less, respectively, than that of L₁. Find the slopes of L₂ and L₃.

Think about what the problem is asking. The angle of inclination for a line is the angle formed between that line and the x-axis. Try drawing a segment of the line L₁. For simplicity's sake I'd pick one of the points as (0,0). Then draw a line segment from the other point on L₁ to the x-axis. Now you have a right triangle. Draw in the angle of inclination. Recall the three basic trig functions (sine, cosine, and tangent). What can you say about that angle (call it theta)?