Answer:
Step-by-step explanation:
Represent the diagonals as:
Required
Determine the coordinate of the intersection
To do this, we simply calculate the midpoint of AC or BD.
For AC:
The midpoint is:
This gives:
For BD:
The midpoint is:
This gives:
Notice the midpoints are the same:
<em>Hence, the coordinates of the intersection is (2,1)</em>
Step-by-step explanation:
the slope of a line is the factor of x, when the equating looks like
y = ...
so,
x - 3y = 3
x = 3y + 3
3y = x - 3
y = 1/3 x - 1
so, the original slope is 1/3.
a parallel line has the same slope.
having a specific point we can use the point-slope form as equation :
y - y1 = m(x - x1)
with m being the slope, and (x1, y1) being a point on the line.
y - -3 = 1/3(x - 2)
y + 3 = 1/3(x - 2)
simplified that is
y + 3 = 1/3 x - 2/3
y = 1/3 x - 2/3 - 3 = 1/3 x - 2/3 - 9/3 = 1/3 x - 11/3 =
= 1/3 × (x - 11)
Well, If there is 16 packs and 8 in each pack then you multiply 16 and 8 which is 128, then add the extra 9 cans of juice and the answer is 137. Have a good rest of your day/night!!
<h3>
<u>Given</u><u> </u><u>:</u><u>-</u></h3>
- PQ = 8cm
- Radius = 5cm
- Two Tangents = P & Q.
<h3>
<u>Construction</u><u> </u><u>:</u><u>-</u></h3>
<h3>
<u>⟼</u><u> </u><u>Solution</u><u> </u><u>:</u><u>-</u></h3>
Here, ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.
[∵ TP=TQ = Tangents from T upon the circle]
⠀⠀⠀⠀⠀⠀⠀⠀∴ OT⊥PQ
⠀⠀⠀
___________________________________________
By Applying Pythagoras Theorem in ∆OPR :
OR = √OP² - PR²
OR = √5² - 4²
OR = 3cm
__________________________________________
Now,
∠TPR + ∠RPO = 90° (∵TPO=90°)
∠TPR + ∠PTR (∵TRP=90°)
<u></u><u>∴ ∠RPO = ∠PTR</u>
⠀⠀
<u>∴ Right triangle TRP is similar to the right </u><u>triangle</u><u> </u><u>PR</u><u>O</u><u>.</u> [By A-A Rule of similar triangles]
⟼
⟼
⟼
<h3>Hence you got your answer here. </h3>
⠀⠀⠀⠀⠀
<h2>-MissAbhi</h2>
